96 Cartesian coordinates; applications Ch. 6
ThusZdivides(Z 1 ,Z 2 )in the ratio|λ|:1.
Changing our notation slightly, if we denote byZ 3 ≡(x 3 ,y 3 )the point with
x 3 =x 1 +λ
1 +λ(x 2 −x 1 )=1
1 +λx 1 +λ
1 +λx 2 ,y 3 =y 1 +λ
1 +λ(y 2 −y 1 )=1
1 +λy 1 +λ
1 +λy 2 ,thenZ 3 divides(Z 1 ,Z 2 )in the ratio|λ|: 1. Consequently if we denote byZ 4 ≡(x 4 ,y 4 )
the point with
x 4 =1
1 +λ′x 1 +λ′
1 +λ′x 2 ,y 4 =1
1 +λ′y 1 +λ′
1 +λ′y 2 ,whereλ′=−λ,sothat
x 4 =1
1 −λx 1 −λ
1 −λ,y 4 =1
1 −λy 1 −λ
1 −λy 2 ,thenZ 4 also divides(Z 1 ,Z 2 )in the ratio|−λ|:1=|λ|:1.
Nowλ= 1 −ttand if we write−λ= 1 −sswe haveZ 4 in the original format,
x 4 =x 1 +s(x 2 −x 1 ),y 4 =y 1 +s(y 2 −y 1 ).Then t
1 −t=−
s
1 −s
so that
s=1
2 t
t−^12.
Thus
s−^12 =1
2 t
t−^12−
1
2
=
1
2 t−1
2 t+1
4
t−^12=^14
1
t−^12.
Hence (
s−^12
)(
t−^12)
=^14 ,
∣∣
(s−^12 )(t−^12 )∣∣
=^14. (6.7.1)
Then we have three possibilities,
(a)∣∣
t−^12∣∣
<^12 ,
∣∣
s−^12∣∣
>^12 ,
(b)∣∣
s−^12∣∣
<^12 ,
∣∣
t−^12∣∣
>^12 ,
(c)∣
∣s−^12∣
∣=^12 ,
∣
∣t−^12∣
∣=^12.
In (a) we have−^12 <t−^12 <^12 and eithers−^12 <−^12 ors−^12 >^12. Hence 0<t< 1
and eithers<0ors>1. It follows thatZ 3 ∈[Z 1 ,Z 2 ],Z 4 ∈[Z 1 ,Z 2 ].
The situation in (b) is like that in (a) with the roles oftands, and so ofZ 3 andZ 4
interchanged.